设y=f(x)由方程2y^3-2y^2+2xy-x^2=1所确定,试求y=f(x)的极值.

设y=f(x)由方程2y^3-2y^2+2xy-x^2=1所确定,试求y=f(x)的极值.
设y=f(x)由方程2y^3-2y^2+2xy-x^2=1所确定,试求y=f(x)的极值.

设y=f(x)由方程2y^3-2y^2+2xy-x^2=1所确定,试求y=f(x)的极值.
求导得：6y²y'-4yy'+2y+2xy'-2x=0
3y²y'-2yy'+y+xy'-x=0
y‘(3y²-2y+x)=x-y
y'=(x-y)/(3y²-2y+x)
极值,则：y'=0,得：x=y
把y=x代入方程：2y³-2y²+2xy-x²=1得：2x³-2x²+2x²-x²=1
2x³-x²=1
x³-x²+x³-1=0
x²(x-1)+(x-1)(x²+x+1)=0
(x-1)(2x²+x+1)=0
得：x=1
则：y=1
所以,极值为1